Question

in the circuit shown, the resistance of the ammeter A is negligible and that of the voltmeter V is very high . when the key K is open, the reading of voltmeter is 1.53 V . on closing the key, the reading of the voltmeter is 1A and that of the voltmeter drops to 1.03 V . calculate : 1 ) emf of the cell 2) internal resistance of the cell 3) value of R 4) percentage energy dissipation in R from the total energy taken from the cell . Hint: percentage energy dissipation in R = i^2 R/ Ei×100. answer 1)1.53 V 2) 0.50 ohm 3)1.03 ohm 4) 67.3% .​

Answer 1

Explanation:

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